Explanation Given: The given Laplace equation is ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0 , 0 < x < a , 0 < y < b and the given boundary conditions are u ( 0 , y ) = 10 y , ∂ u ∂ x | x = 1 = − 1 , u ( x , 0 ) = 0 and u ( x , 1 ) = 0 . Calculation: The given Laplace equation for rectangular plate is, ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0 Consider the equation, u ( x , y ) = X ( x ) Y ( y ) (1) Separate the variables of the above equation, X ″ X = − λ and − Y ″ Y = − λ The equation of the variable X is, X ″ + λ X = 0 (2) The equation of the variable Y is, Y ″ − λ Y = 0 (3) The solution of the equations can be obtained for various cases considering the value of λ to be strictly positive, strictly negative and zero. When λ = 0 , Substitute λ = 0 in equation (2), X ″ = 0 The general solution of the above equation is, X ( x ) = c 1 x + c 2 (4) At boundary conditions ∂ u ∂ x | x = 1 = − 1 , u ( x , 0 ) = 0 and u ( x , 1 ) = 0 , c 1 = − 1 and c 2 = x Substitute c 1 = − 1 and c 2 = x in the equation (4), X ( x ) = 0 (5) Substitute λ = 0 in equation (3), Y ″ = 0 The general solution of the above equation is, Y ( y ) = c 3 y + c 4 (6) At boundary conditions ∂ u ∂ x | x = 1 = − 1 , u ( x , 0 ) = 0 and u ( x , 1 ) = 0 , c 3 = 0 and c 4 = 0 Substitute c 3 = 0 and c 4 = 0 in the equation (6), Y ( y ) = 0 (7) Substitute the value of X ( x ) from equation (5) and the value of Y ( y ) from equation (7) in equation (1), u ( x , y ) = 0 When λ = − α 2 , Substitute λ = − α 2 in equation (2), X ″ − α 2 X = 0 The general solution of the above equation is, X ( x ) = c 1 cosh α x + c 2 sinh α x (8) Substitute λ = − α 2 in equation (3), Y ″ + α 2 Y = 0 The general solution of the above equation is, Y ( y ) = c 3 cos α y + c 4 sin α y (9) At boundary conditions ∂ u ∂ x | x = 1 = − 1 , u ( x , 0 ) = 0 and u ( x , 1 ) = 0 , c 3

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Chapter 12.5, Problem 10E

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Chapter 12.5, Problem 11E

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Problem 11 (a) A tank is discharging water through an orifice at a depth of T meter below the surface of the water whose area is A m². The following are the values of a for the corresponding values of A: A 1.257 1.390 x 1.50 1.65 1.520 1.650 1.809 1.962 2.123 2.295 2.462|2.650 1.80 1.95 2.10 2.25 2.40 2.55 2.70 2.85 Using the formula -3.0 (0.018)T = dx. calculate T, the time in seconds for the level of the water to drop from 3.0 m to 1.5 m above the orifice. (b) The velocity of a train which starts from rest is given by the fol- lowing table, the time being reckoned in minutes from the start and the speed in km/hour: | † (minutes) |2|4 6 8 10 12 14 16 18 20 v (km/hr) 16 28.8 40 46.4 51.2 32.0 17.6 8 3.2 0 Estimate approximately the total distance ran in 20 minutes.

- Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p − 1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., p-1 2 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). 23 32 how come? The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. The set T is the subset of these residues exceeding So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1.

Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p-1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., 2 p-1 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. 23 The set T is the subset of these residues exceeding 2° So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1. how come?

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